Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1);
const int INF = 1e9;
const int N = 1e4 + 1;
const int LOG = 30;
vector<int> inds[N];
vector<pair<int, int> > answer[2];
int a[N], b[N];
void f(int* a, int i, int j, int ai) {
  answer[ai].push_back({i, j});
  a[i] ^= a[j];
}
vector<int> gauss(int* a, int n, int ai) {
  vector<int> bits;
  int cntFound = 0;
  for (int lg = LOG - 1; lg >= 0; lg--) {
    int pos = -1;
    for (int i = cntFound; i < n; i++) {
      if (a[i] & (1 << lg)) {
        pos = i;
        break;
      }
    }
    if (pos != -1) {
      bits.push_back(lg);
      if (cntFound != pos) {
        f(a, cntFound, pos, ai);
        f(a, pos, cntFound, ai);
        f(a, cntFound, pos, ai);
      }
      for (int i = 0; i < n; i++)
        if (cntFound != i && (a[i] & (1 << lg))) {
          f(a, i, cntFound, ai);
        }
      cntFound++;
    }
  }
  return bits;
}
int a_r[N], b_r[N];
void solve() {
  int n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a_r[i] = a[i];
  }
  auto bitsa = gauss(a, n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    b_r[i] = b[i];
  }
  auto bitsb = gauss(b, n, 1);
  if (bitsb.size() > bitsa.size()) {
    cout << -1 << "\n";
    return;
  }
  for (int i = 0; i < bitsb.size(); i++) {
    if (!(b[i] & (1 << bitsa[i]))) {
      f(a, i, i, 0);
    }
    for (int j = i + 1; j < bitsa.size(); j++) {
      if (b[i] & (1 << bitsa[j])) {
        f(a, i, j, 0);
      }
    }
  }
  for (int i = (int)bitsb.size(); i < (int)bitsa.size(); i++) {
    f(a, i, i, 0);
  }
  reverse(answer[1].begin(), answer[1].end());
  for (auto x : answer[1]) answer[0].push_back(x);
  for (auto x : answer[0]) {
    a_r[x.first] ^= a_r[x.second];
  }
  for (int i = 0; i < n; i++)
    if (a_r[i] != b_r[i]) {
      cout << "-1\n";
      return;
    }
  cout << answer[0].size() << "\n";
  for (auto x : answer[0]) cout << x.first + 1 << " " << x.second + 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}